zoj - 2412 - Farm Irrigation
2012-11-10 00:47
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连块问题,输入矩阵后,逐点扫描,该点若没未标记过,计数器就+1,然后深度优先遍历,标记这一大块的通路,接着,继续扫点。
#include <iostream> #include <string.h> using namespace std; const int maxn = 50 + 10; //设置最大长度 struct Tmap //定义结点数据类型,共有11种块,故开一个大小为12的数组,加1个四个方向都没有的结点 { int up; //结点的上面有没通路,有的话为1,没有为0 int down; //结点的下面有没通路,有的话为1,没有为0 int left; //结点的左边有没通路,有的话为1,没有为0 int right; //结点的右边有没通路,有的话为1,没有为0 }map[12]; int M, N, st[maxn][maxn], vis[maxn][maxn]; //M、N为输入的行数和列数,st为转换后的整型数组,vis为状态数组 void dfs(int x, int y) //深度优先遍历,将属于同一块的都标记掉 { if(vis[x][y] == 1) //如果该点已经访问过,返回 return; vis[x][y] = 1; //否则,修改状态 if(x > 1 && map[st[x-1][y]].down*map[st[x][y]].up == 1) dfs(x-1, y); //判断可否向上走 if(y < N && map[st[x][y+1]].left*map[st[x][y]].right == 1) dfs(x, y+1); //判断可否向右走 if(x < M && map[st[x+1][y]].up*map[st[x][y]].down == 1) dfs(x+1, y); //判断可否向下走 if(y > 1 && map[st[x][y-1]].right*map[st[x][y]].left == 1) dfs(x, y-1); //判断可否向左走 } int main() { map[0].up = 1; map[0].right = 0; map[0].down = 0; map[0].left = 1; //A map[1].up = 1; map[1].right = 1; map[1].down = 0; map[1].left = 0; //B map[2].up = 0; map[2].right = 0; map[2].down = 1; map[2].left = 1; //C map[3].up = 0; map[3].right = 1; map[3].down = 1; map[3].left = 0; //D map[4].up = 1; map[4].right = 0; map[4].down = 1; map[4].left = 0; //E map[5].up = 0; map[5].right = 1; map[5].down = 0; map[5].left = 1; //F map[6].up = 1; map[6].right = 1; map[6].down = 0; map[6].left = 1; //G map[7].up = 1; map[7].right = 0; map[7].down = 1; map[7].left = 1; //H map[8].up = 0; map[8].right = 1; map[8].down = 1; map[8].left = 1; //I map[9].up = 1; map[9].right = 1; map[9].down = 1; map[9].left = 0; //J map[10].up = 1; map[10].right = 1; map[10].down = 1; map[10].left = 1; //K int i, j; char s[maxn][maxn]; //输入的字符型数组 while(cin>>M>>N) { if(M < 0 || N < 0) return 0; //如果有一个数小于0,返回 memset(vis, 0, sizeof(vis)); //初始化所有点为0,可以访问 for(i = 1; i <= M; i++) { for(j = 1; j <= N; j++) { cin>>s[i][j]; st[i][j] = s[i][j] - 'A'; //输入后即转化为整型,A、B~K分别对应0、1~10 } } for(i = 0; i <= N+1; i++) //外加一层边,置其所有走向为0(只需让其成为map[11]) { st[0][i] = 11; st[M+1][i] = 11; } for(i = 0; i <= M+1; i++) { st[i][0] = 11; st[i][N+1] = 11; } int cnt = 0; //计数器 for(i = 1; i <= M; i++) //每个点扫一次 for(j = 1; j <= N; j++) { if(vis[i][j] == 0) //如果没有走过,就加1块,并把这一大块标记掉 { cnt++; dfs(i, j); //它的功能就是标记掉这一大块 } } cout<<cnt<<endl; } return 0; }
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